Determine a set of all vectors of $V$ that have the same coordinates in different bases

Question

Let $B=(b_1,b_2,b_3)$ and $C=(c_1,c_2,c_3)$ are two different bases of three dimensional vector space $V$ such that: $$c_1=b_1+3b_2+3b_3$$ $$c_2=4b_2+5b_3$$ $$c_3=2b_3$$

Determine a set of all vectors of $V$ that have the same coordinates in both bases. Check if that set is a vector space and if it is, find its dimension and one basis.

Question: Problem doesn't state what are dimensions of vectors $b_1,b_2,b_3,c_1,c_2,c_3$. How can we find dimension and one basis of a vector space if we don't know the vectors in explicit form?

Answer 1

Each such vector $x\in V$ corresponds to a number triple $(x_1,x_2,x_3)$ satisfying $$(x=)\quad x_1c_1+x_2c_2+x_3c_3=x_1b_1+x_2b_2+x_3b_3\ .$$ By definition of the $c$-basis $$\eqalign{LHS&=x_1(b_1+3b_2+3b_3)+x_2(4b_2+5b_3)+x_3(2b_3)\cr&=x_1b_1+(3x_1+4x_2)b_2+(3x_1+5x_2+2x_3)b_3\ .\cr}$$ Since $(b_1,b_2,b_3)$ is a basis of $V$ this implies that we have the three equations $$\eqalign{x_1&=x_1\cr3x_1+4x_2&=x_2\cr3x_1+5x_2+2x_3&=x_3\ .\cr}$$ This can be recursively solved as follows: $$x_1={\rm arbitrary},\quad x_2=-x_1,\quad x_3=2x_1\ .$$ The solution set in the triples space ${\mathbb R}^3$ is therefore the set of all triples $\lambda(1,-1,2)$, $\lambda\in{\mathbb R}$, and the set $S$ of all vectors $x\in V$ satisfying the given condition is given by $$S=\{\lambda b_*\,|\,\lambda\in{\mathbb R}\},\qquad b_*=b_1-b_2+2b_3\ ,$$ and is a one-dimensional vector space with basis vector $b_*$.

Answer 2

When you find that $\\(x_1,x_2,x_3)=x_2(-1,1,-2)$ you will know that all the vectors in S can be represented as linear combination of $\{-b_1+b_2-2b_3\} or \{-c_1+c_2-2c_3\} $ and the set of all linear combinations is a span over those vectors. Span is a subspace.

Answer 3

What are the "dimensions" of a vector? If you mean the coordinates of the vector, then components are only defined relative to a basis.

You have been given a basis of $V,$ therefore you can express any vector as a list of coordinates over that basis. Actually you have been given two bases, but that just gives you a second way to give coordinates of a vector; it does not prevent you from using the first basis you were given.

So you have $b_1,$ $b_2,$ and $b_3,$ which are a basis. (This is given.) You can express any other basis of the space $V,$ or the basis of any subspace of $V,$ in terms of these three vectors. One such example has already been given in the problem statement: the three vectors $b_1+3b_2+3b_3,$ $4b_2+5b_3,$ and $2b_3$ (also known as $c_1,$ $c_2,$ and $c_3$) are a basis of $V.$ This is a perfectly valid way to write a basis of a vector space, given that we know the vectors $b_1,$ $b_2,$ and $b_3$ are also a basis, and it is the way this question apparently is meant to be answered.

For example, consider the two vectors $b_1$ and $b_1 + b_2 + b_3.$ These vectors are independent, since otherwise you would be able to show that $b_2 + b_3 = 0.$ The span of these two vectors therefore has two dimensions, and because both vectors are vectors in $V,$ their span is a two-dimensional subspace of $V,$ namely, $\{t_1b_1 + t_2(b_1+b_2+b_3)\}.$ The same two vectors ($b_1$ and $b_1 + b_2 + b_3$) are a basis of that subspace.

In case you have not already gotten the hint from this example, when you wrote $S=\{x_2(-b_1+b_2-2b_3)\}$ you had already practically solved the problem. You just need to read off the dimension and the basis from what you wrote.

You do not need to find "coordinates" of $b_1$ or of any of the other given basis vectors of $V$ to solve this problem. The only meaningful coordinates for $b_1$ other than $(1,0,0)$ (defined in terms of the basis $(b_1,b_2,b_3)$) or $\left(1,-\frac34,\frac38\right)$ (defined in terms of the basis $(c_1,c_2,c_3)$) would be defined in terms of some third basis that has not been given--and no such basis has any better claim to be your chosen basis than either $(b_1,b_2,b_3)$ or $(c_1,c_2,c_3).$ So don't look for something you don't have and don't need.